Question

Help me
Find Lim sin(2-cos(x) - cos (2x) ) / x and x approaches zero

Answer 1

Hey there :)

Here is an helpful limit to recall:\[\large\rm \color{orangered}{\lim_{x\to0}\frac{\sin x}{x}=1}\]

Answer 2

It might help us here, hmm thinking :d

Answer 3

Hmm I can't find a normal approach to this one.
Are you familiar with L'Hospital's Rule?

Answer 4

No

Answer 5

@Michele_Laino @Empty @AlexandervonHumboldt2 hmm any ideas people? :)

Answer 6

we can use this identity:
\[\begin{gathered}
\frac{{\sin \left( {2 - \cos x - \cos \left( {2x} \right)} \right)}}{x} = \hfill \\
\hfill \\
= \frac{{\sin \left( {2 - \cos x - \cos \left( {2x} \right)} \right)}}{{2 - \cos x - \cos \left( {2x} \right)}} \cdot \frac{{2 - \cos x - \cos \left( {2x} \right)}}{x} = \hfill \\
\hfill \\
= \frac{{\sin \left( {2 - \cos x - \cos \left( {2x} \right)} \right)}}{{2 - \cos x - \cos \left( {2x} \right)}} \cdot \frac{{1 - \cos x + 1 - \cos \left( {2x} \right)}}{x} = \hfill \\
\hfill \\
= \frac{{\sin \left( {2 - \cos x - \cos \left( {2x} \right)} \right)}}{{2 - \cos x - \cos \left( {2x} \right)}} \cdot \left\{ {\frac{{1 - \cos x}}{x} + \frac{{2\left( {1 - \cos \left( {2x} \right)} \right)}}{{2x}}} \right\} = \hfill \\
\hfill \\
= \frac{{\sin \left( {2 - \cos x - \cos \left( {2x} \right)} \right)}}{{2 - \cos x - \cos \left( {2x} \right)}} \cdot \left\{ {\frac{{2{{\left( {\sin \left( {x/2} \right)} \right)}^2}}}{x} + \frac{{4{{\left( {\sin x} \right)}^2}}}{x}} \right\} \hfill \\
\end{gathered} \]

Answer 7

Looks like we're using the chain rule

Answer 8

Thank you 😊